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Chapter 11: Problem 46

The tension in a guitar string is increased by \(15 \% .\) What happens to the fundamental frequency of the string?

Short Answer

Expert verified
The fundamental frequency increases by approximately 7.2%.

Step by step solution

01

Understand the Relationship

The fundamental frequency \( f \) of a string is given by the formula \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension, \( L \) is the length, and \( \mu \) is the linear mass density of the string. We can see that the frequency depends on the square root of the tension \( T \).
02

Calculate the Effect of Tension Increase

If the tension is increased by 15%, the new tension becomes \( T_{\text{new}} = T + 0.15T = 1.15T \). We want to find the new frequency \( f_{\text{new}} \).
03

Develop the New Frequency Equation

Substitute the new tension into the frequency formula: \[ f_{\text{new}} = \frac{1}{2L} \sqrt{\frac{1.15T}{\mu}}. \]
04

Simplify the Expression

Factor out the change in tension from the square root:\[ f_{\text{new}} = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \times \sqrt{1.15}. \]Since \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), we have \( f_{\text{new}} = f \times \sqrt{1.15}. \)
05

Find the Numerical Increase in Frequency

Calculate \( \sqrt{1.15} \approx 1.072 \). This indicates that the new frequency is approximately 1.072 times the original frequency.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fundamental Frequency
The fundamental frequency, often just called the first harmonic, is the lowest frequency at which a string vibrates when it is plucked or bowed. It is the most basic tone produced by any vibrating object, such as a guitar string. For a string instrument, this frequency is crucial as it defines the pitch of the note being played.
  • The fundamental frequency, denoted by \( f \), is determined by various factors, including the tension in the string, the length of the string, and the linear mass density of the string material.
  • The mathematical representation is given by \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), where \( L \) is the length of the string, \( T \) is the tension, and \( \mu \) is the mass per unit length of the string.
Understanding this formula helps us see how changes in one of these variables can affect the fundamental frequency. An increase in any of the factors related to tension or decrease in linear mass density will result in an increase in fundamental frequency.
Tension and Frequency Relationship
The tension of a string has a direct impact on the frequency of the sound it produces. Intuitively, if you stretch a guitar string tighter, it vibrates faster, leading to a higher pitch.
  • As per the fundamental frequency formula, \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), the frequency \( f \) is proportional to the square root of the tension \( T \).
  • Therefore, if tension increases, the frequency also increases, specifically by the square root of the fractional increase in tension.
  • For example, when the tension is increased by 15%, this can be represented as \( T_{\text{new}} = 1.15T \). The new frequency, \( f_{\text{new}} = f \times \sqrt{1.15} \), shows that frequency changes by a factor of the square root of the increase.
This principle is why musicians precisely adjust the tension of their instrument strings to accurately tune to the desired pitch.
String Instruments Physics
String instruments, like guitars and violins, rely heavily on vibrations to produce music. These vibrations are inherently physical phenomena subject to the laws of physics.
  • When a string is plucked, it vibrates back and forth, setting air molecules in motion to create sound waves.
  • The vibration occurs in segments: the entire length of the string vibrates for the fundamental frequency, while shorter segments of the string create overtones or harmonics.
  • Factors that affect the vibration include tension, length, and linear mass density, along with environmental factors like temperature and humidity.
Understanding these physical properties allows musicians and instrument designers to manipulate and optimize conditions to achieve the desired sound quality. This understanding highlights the beautiful intersection of art and science present in string instruments.

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Most popular questions from this chapter

(a) Use a graphing calculator or computer graphing program to plot \(y\) versus \(x\) for the function $$ y(x, t)=(5.0 \mathrm{cm})[\sin (k x-\omega t)+\sin (k x+\omega t)] $$ for the times \(t=0,1.0 \mathrm{s},\) and \(2.0 \mathrm{s}\). Use the values \(k=\pi /(5.0 \mathrm{cm})\) and \(\omega=(\pi / 6.0) \mathrm{rad} / \mathrm{s} .\) (b) Is this a traveling wave? If not, what kind of wave is it?
Using graph paper, sketch two identical sine waves of amplitude $4.0 \mathrm{cm}\( that differ in phase by (a) \)\pi / 3$ rad \(\left(60^{\circ}\right)\) and (b) \(\pi / 2\) rad \(\left(90^{\circ}\right) .\) Find the amplitude of the superposition of the two waves in each case.
Interference and Diffraction Two waves with identical frequency but different amplitudes $A_{1}=5.0 \mathrm{cm}\( and \)A_{2}=3.0 \mathrm{cm},$ occupy the same region of space (are superimposed). (a) At what phase difference does the resulting wave have the largest amplitude? What is the amplitude of the resulting wave in that case? (b) At what phase difference does the resulting wave have the smallest amplitude and what is its amplitude? (c) What is the ratio of the largest and smallest amplitudes?
A fisherman notices a buoy bobbing up and down in the water in ripples produced by waves from a passing speedboat. These waves travel at $2.5 \mathrm{m} / \mathrm{s}\( and have a wavelength of \)7.5 \mathrm{m} .$ At what frequency does the buoy bob up and down?
Reflection and Refraction Light of wavelength \(0.500 \mu \mathrm{m}\) (in air) enters the water in a swimming pool. The speed of light in water is 0.750 times the speed in air. What is the wavelength of the light in water?
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